Multipliers and weak multipliers of algebras

Abstract

We investigate general properties of multipliers and weak multipliers of algebras. We apply the results to determine the (weak) multipliers of associative algebras and zeropotent algebras of dimension 3 over an algebraically closed field.

1 Introduction

Multipliers of algebras, particularly multipliers of Banach algebras, have been studied in the field of analysis. In this paper we discuss them in a purely algebraic manner.

Let A be a Banach algebra. A mapping $T: A \rightarrow A$ is termed a multiplier of A if it satisfies the condition (I) $xT(y) = T(xy) = T(x)y$ for all $x, y \in A$. We denote the collection of all multipliers of A as M(A), and the collection of all bounded linear operators on A as B(A). Notably, M(A) forms an algebra and B(A) constitutes a Banach algebra. A Banach algebra A is referred to as without order if it has neither a nonzero left annihilator nor a nonzero right annihilator. If A is without order, the algebra M(A) forms a commutative closed subalgebra of B(A) (see [2], Proposition 1.4.11). In 1952, Wendel [8] proved an important result that the multiplier algebra of $L^1(G)$ on a locally compact group G is isometrically isomorphic to the measure algebra on G. The general theory of multipliers of Banach algebras has been developed by Johnson [1]. For a comprehensive reference on the theory of multipliers of Banach algebras, refer to Larsen [5].

When A is without order, T is a multiplier if it satisfies the condition (II) $xT(y) = T(x)y$ for all $x, y \in A$. Many researchers had been unaware of the difference between conditions (I) and (II) until Zivari-Kazempour [9] (see also [10]) recently articulated the difference. We call a mapping T satisfying (II) a weak multiplier and denote the set of such multipliers of A by $M'(A)$. Then, M(A) is in general a proper subset of $M'(A)$. Furthermore, (weak) multipliers can be defined for an algebra A that is not necessarily associative, and they are not linear mappings in general. We denote the spaces of linear multipliers and linear weak multipliers of A by LM(A) and $LM'(A)$ respectively. M(A) and LM(A) are subalgebras of the algebra $A^A$ consisting of all mappings from A to itself. Meanwhile, $M'(A)$ and $LM'(A)$ are closed under the operation $\circ$ defined by $T \circ S = TS + ST$, and they form a Jordan algebra.

In Sects. 2 to 5, we study general properties of (weak) multipliers. More specifically, in Sects. 3 and 4 we give a decomposition theorem (Theorem 3.1), and a matrix equation (Theorem 4.2) for (weak) multipliers. They play an essential role in our examination of (weak) multipliers.

Complete classifications of associative algebras and zeropotent algebras of dimension 3 over an algebraically closed field of characteristic not equal to 2 were given in Kobayashi et al., [3] and [4]. In Sects. 6 and 7 we undertake a complete determination of the (linear) (weak) multipliers of these algebras.

Some authors have considered other weaker concepts related to multipliers, such as (pseudo-)n-multipliers (for more information, see [6] and [11]).

2 Multipliers and weak multipliers

Let K be a field and A be a (not necessarily associative) algebra over K. The set $A^A$ of all mappings from A to A forms an associative algebra over K in the usual way. Let L(A) denote the subalgebra of $A^A$ of all linear mappings from A to A.

A mapping $T: A \rightarrow A$ is a weak multiplier of A, if

$$\begin{aligned} xT(y) = T(x)y \end{aligned}$$

(1)

holds for any $x, y \in A$, and T is a multiplier, if

$$\begin{aligned} xT(y) = T(xy) = T(x)y \end{aligned}$$

(2)

for any $x, y \in A$. Let M(A) (resp. $M'(A)$) denote the set of all multipliers (resp. weak multipliers) of A. Define

$$\begin{aligned} LM(A) \overset{\text {\tiny def}}{=}M(A) \cap L(A)\ \, \text{ and }\ \, LM'(A) \overset{\text {\tiny def}}{=}M'(A) \cap L(A). \end{aligned}$$

Proposition 2.1

M(A) (resp. LM(A)) is a unital subalgebra of $A^A$ (resp. L(A)), and $M'(A)$ (resp. $LM'(A)$) is a Jordan subalgebra of $A^A$ (resp. L(A)).

Proof

First, the zero mapping is a multiplier because all three terms in (2) are zero. Secondly, the identity mapping is also a multiplier because the three terms in (2) are xy. Let $T, U \in M(A)$. Then we have

$$\begin{aligned} x(T+U)(y)= & {} xT(y) + xU(y) = T(xy) + U(xy) = T(x)y + U(x)y\nonumber \\= & {} (T+U)(x)y \end{aligned}$$

(3)

and

$$\begin{aligned} x(TU)(y) = xT(U(y)) = T(xU(y)) = TU(xy) = T( U(x)y) = (TU)(x)y\nonumber \\ \end{aligned}$$

(4)

for any $x, y \in A$. Hence, $T+U, TU$ belong to M(A). Finally let $k \in K$, then

$$\begin{aligned} x(kT)(y) = kxT(y) = kT(xy) = kT(x)y = (kT)(x)y, \end{aligned}$$

(5)

and so $kT \in M(A)$. Therefore, M(A) is a unital subalgebra of $A^A$, and $LM(A) = M(A)\cap L(A)$ is a unital subalgebra of L(A).

Next, let $T, U \in M'(A)$. Then, the equalities in (3) and (5) hold, removing the center terms $T(xy)+U(xy)$ and kT(xy), respectively. Hence, $M'(A)$ is a subspace of $A^A$. Moreover, we have

$$\begin{aligned} x(TU)(y) = xT(U(y)) = T(x)U(y) = U(T(x))y =UT(x)y \end{aligned}$$

and similarly $x(UT)(y) = TU(x)y$ for any $x, y \in A$. Hence,

$$\begin{aligned} x(TU + UT)(y) = (TU + UT)(x)y. \end{aligned}$$

It follows that $TU + UT \in M'(A)$.¹$\square$

Let Ann$_l(A)$ (resp. Ann$_r(A)$) be the left (resp. right) annihilator of A and let $A_0$ be their intersection, that is,

$$\begin{aligned} \textrm{Ann}_l(A) = \{a \in A\,|\, ax = 0 {\ \mathrm for\ all\ } x \in A\}, \\ \textrm{Ann}_r(A) = \{a \in A\,|\, xa = 0 {\ \mathrm for\ all\ } x \in A\} \end{aligned}$$

and

$$\begin{aligned} A_0 = \textrm{Ann}_l(A) \cap \textrm{Ann}_r(A). \end{aligned}$$

They are all subspaces of A, and when A is an associative algebra, they are two-sided ideals. For a subset X of A, $\langle X\rangle$ denotes the subspace of A generated by X.

Proposition 2.2

A weak multiplier T of A such that $\langle T(A)\rangle \cap A_0 = \{0\}$ is a linear mapping.

Proof

Let $x, y, z \in A$ and $a, b \in K$, and let T be a weak multiplier. We have

$$\begin{aligned} T(ax + by)z&= (ax+by)T(z) = axT(z) + byT(z) = aT(x)z + bT(y)z\\&= \left( aT(x) + bT(y)\right) z. \end{aligned}$$

Because z is arbitrary, we have $w = T(ax+by) - aT(x) - bT(y) \in \textrm{Ann}_l(A)$. Similarly, we can show $w \in \textrm{Ann}_r(A)$, and so $w \in A_0$. Hence, if $\langle T(A)\rangle \cap A_0 = \{0\}$, then $w = 0$ because $w \in \langle T(A)\rangle$. Since a, b, x, y are arbitrary, T is a linear mapping. $\square$

Corollary 2.3

If $A_0 = \{0\}$, then any weak multiplier is a linear mapping over K, that is, $M'(A) = LM'(A)$ and $M(A) = LM(A)$.

Proposition 2.4

If T is a weak multiplier, then $T\left( \textrm{Ann}_l(A)\right) \subseteq \textrm{Ann}_l(A)$, $T\left( \textrm{Ann}_r(A)\right) \subseteq \textrm{Ann}_r(A)$ and $T(A_0) \subseteq A_0$.

Proof

Let $x \in \textrm{Ann}_l(A)$, then for any $y \in A$ we have

$$\begin{aligned} 0 = xT(y) = T(x)y. \end{aligned}$$

Hence, $T(x) \in \textrm{Ann}_l(A)$. The other cases are similar. $\square$

In this paper we denote the subset $\{xy\,|\,x, y \in A\}$ of A by $A^2$.²

Proposition 2.5

Any mapping $T\hspace{-.6mm}: A \rightarrow A$ such that $T(A) \subseteq A_0$ is a weak multiplier. Such a mapping T is a multiplier if and only if $T(A^2) = \{0\}$. In particular, if A is the zero algebra, every mapping T is a weak multiplier, and it is a multiplier if and only if $T(0) = 0$.

Proof

If $T(A) \subseteq A_0$, then both sides are 0 in (1) and T is a weak multiplier. This T is a multiplier if and only if the term T(xy) in the middle of (2) is 0 for all $x, y \in A$, that is, $T(A^2) = \{0\}$. If A is the zero algebra, then $A = A_0$ and $A^2$ = {0}. Hence, any T is a weak multiplier and it is a multiplier if and only if $T(0) = 0$. $\square$

The opposite $A^\textrm{op}$ of A is the algebra with the same elements and addition as A, but the multiplication $*$ in it is reversed, that is, $x*y = yx$ for all $x, y \in A$.

Proposition 2.6

A and $A^\textrm{op}$ have the same multipliers and weak multiplies, that is,

$$\begin{aligned} M(A^\textrm{op}) = M(A)\ \ \text{ and }\ \ M'(A^\textrm{op}) = M'(A). \end{aligned}$$

Proof

Let $T \in A^A$. Then, $T \in M'(A)$ if and only if

$$\begin{aligned} x*T(y) = T(y)x = yT(x) = T(x)*y \end{aligned}$$

for any $x, y \in A$, if and only if $T \in M'(A^\textrm{op})$. Further, $T \in M(A)$ if and only if

$$\begin{aligned} x*T(y) = T(y)x = T(yx) = T(x*y) = yT(x) = T(x)*y \end{aligned}$$

for any $x, y \in A$, if and only if $T \in M(A^\textrm{op})$. $\square$

3 Nihil decomposition

Let $A_1$ be a subspace of A such that

$$\begin{aligned} A = A_1 \oplus A_0. \end{aligned}$$

(6)

Here, $A_1$ is not unique, but choosing an appropriate $A_1$ will become important later. When $A_1$ is fixed, any mapping $T \in A^A$ is uniquely decomposed as

$$\begin{aligned} T = T_1 + T_0 \end{aligned}$$

(7)

with $T_1(A) \subseteq A_1$ and $T_0(A) \subseteq A_0$. We call (6) and (7) a nihil decomposition of A and T, respectively. Let $\pi : A \rightarrow A_1$ be the projection and $\mu : A_1 \rightarrow A$ be the embedding, that is, $\pi (x_1+x_0) = \mu (x_1) = x_1$ for $x_1 \in A_1$ and $x_0 \in A_0$.

Let $M_1(A)$ (resp. $M_0(A)$) denote the set of all multipliers T of A with $T(A) \subseteq A_1$ (resp. $T(A) \subseteq A_0$). Similarly, the sets $M'_1(A)$ and $M'_0(A)$ of weak multipliers of A are defined. Also, set

$$\begin{aligned} LM_i(A) = M_i(A) \cap L(A)\ \,\text{ and }\ \,LM'_i(A) = M'_i(A) \cap L(A) \end{aligned}$$

for $i = 0,1$. By Proposition 2.2 we see

$$\begin{aligned} M'_1(A) = LM'_1(A)\ \,\text{ and }\ \,M_1(A) = LM_1(A), \end{aligned}$$

and by Proposition 2.5 we have

$$\begin{aligned} M'_0(A) = A_0^A\,\ \text{ and }\,\ M_0(A) = \{T \in A_0^A\,|\,T(A^2) = \{0\}\}. \end{aligned}$$

(8)

Theorem 3.1

Let A = $A_1 \oplus A_0$ and $T = T_1 + T_0$ be nihil decompositions of A and $T \in A^A$ respectively.

(i) T is a weak multiplier, if and only if $T_1$ is a weak multiplier. If T is a weak multiplier, $T_1$ is a linear mapping satisfying $T_1(A_0) = \{0\}$.

(ii) If $T_1$ is a multiplier and $T_0(A^2) = \{0\}$, then T is a multiplier. If $A_1$ is a subalgebra of A, the converse is also true.

Suppose that $A_1$ is a subalgebra of A, and let $\Phi$ be a mapping sending $R \in (A_1)^{A_1}$ to $\mu \circ R \circ \pi \in A^A$. Then,

(iii) $\Phi$ gives an algebra isomorphism from $M(A_1)$ onto $M_1(A)$ and a Jordan isomorphism from $M'(A_1)$ onto $M'_1(A)$.

Proof

Let $x, y \in A$.

(i) If T is a weak multiplier, then

$$\begin{aligned} xT_1(y) = x(T(y) - T_0(y)) = xT(y) = T(x)y = T_1(x)y. \end{aligned}$$

Thus, $T_1$ is also a weak multiplier. Moreover, $T_1$ is a linear mapping by Proposition 2.2 and $T_1(A_0) \subseteq A_1 \cap A_0 = \{0\}$ by Proposition 2.4. Conversely, if $T_1$ is a weak multiplier, then

$$\begin{aligned} xT(y) = xT_1(y) = T_1(x)y = T(x)y, \end{aligned}$$

and so T is a weak multiplier.

(ii) If $T_1$ is a multiplier and $T_0(A^2) = 0$, then T is a multiplier because

$$\begin{aligned} xT(y) = xT_1(y) = T_1(xy) = T(xy) - T_0(xy) = T(xy) = T(x)y. \end{aligned}$$

Next suppose that $A_1$ is a subalgebra. If T is a multiplier, then for any $x, y \in A$ we have

$$\begin{aligned} T_1(xy) + T_0(xy) = T(xy) = xT(y) = x_1T_1(y), \end{aligned}$$

(9)

where $x = x_1+x_0$ with $x_1 \in A_1$ and $x_0 \in A_0$. Here, $x_1T_1(y) \in A_1$ because $A_1$ is a subalgebra, and thus, we have $T_0(xy) = x_1T_1(y) - T_1(xy) \in A_0 \cap A_1 = \{0\}$. Since x, y are arbitrary, we get $T_0(A^2) = \{0\}$. Moreover, because $T_1(xy) = x_1T_1(y) = xT_1(y)$ by (9) and similarly $T_1(xy) = T_1(x)y$, $T_1$ is a multiplier. The converse is already proved above.

(iii) Let $S \in (A_1)^{A_1}$ and $x = x_1+x_0, y = y_1 + y_0 \in A$ with $x_1, y_1 \in A_1$ and $x_0, y_0 \in A_0$. Then, $\pi (x) = \mu (x_1) = x_1$, $\pi (y) = \mu (y_1) = y_1$ and

$$\begin{aligned} \Phi (S)(x) = \mu (S(\pi (x))) = \mu (S(x_1)) = S(x_1). \end{aligned}$$

Thus, if $S \in M'(A_1)$, we have

$$\begin{aligned} x\Phi (S)(y) = xS(y_1) = x_1S(y_1) = S(x_1)y_1 = \Phi (S)(x)y_1 = \Phi (S)(x)y. \end{aligned}$$

Hence, $\Phi (S) \in M'_1(A)$. Moreover, if $S \in M(A_1)$, then because $\pi$ is a homomorphism, we have

$$\begin{aligned} \Phi (S)(xy) = S(\pi (xy)) = S(x_1y_1) = x_1S(y_1) = x\Phi (S)(y), \end{aligned}$$

and so $\Phi (S) \in M_1(A)$.

Conversely, let $T \in M'_1(A)$, then because T is a linear mapping satisfying $T(A_0) = \{0\}$, there is a linear mapping $S \in L(A_1)$ on $A_1$ such that $\Phi (S) = T$, that is, $S(x_1) = T(x) = T(x_1)$. We have

$$\begin{aligned} x_1S(y_1) = x_1T(y_1) = T(x_1)y_1 = S(x_1)y_1, \end{aligned}$$

and hence $S \in M'(A_1)$. Therefore, $\Phi$ induces a linear isomorphism from $M'(A_1)$ to $M'_1(A)$. Similarly, $\Phi$ gives a linear isomorphism from $M(A_1)$ to $M_1(A)$. Moreover, for $T, U \in M'(A_1)$, we have

$$\begin{aligned} \Phi (TU) = \mu \circ T\circ U\circ \pi = \mu \circ T\circ \pi \circ \mu \circ U\circ \pi = \Phi (T)\Phi (U). \end{aligned}$$

Thus, $\Phi$ gives an isomorphism of algebras from $M(A_1)$ to $M_1(A)$ and a Jordan isomorphism from $M'(A_1)$ to $M'_1(A)$. $\square$

Theorem 3.1 implies

$$\begin{aligned} M'(A) = \ M'_1(A) \oplus M'_0(A)\ \text{ and }\ M_1(A) \oplus M_0(A) \subseteq M(A), \end{aligned}$$

where $M'_0(A)$ and $M_0(A)$ are given as (8). If $A_1$ is a subalgebra, we have

$$\begin{aligned} M'(A) \cong M'(A_1) \oplus (A_0)^A\ \text{ and }\ M(A) \cong M(A_1) \oplus \left\{ T \in (A_0)^A\,|\,T(A^2) = \{0\}\right\} .\nonumber \\ \end{aligned}$$

(10)

Corollary 3.2

Any weak multiplier T is written as

$$\begin{aligned} T = T_1 + R \end{aligned}$$

(11)

with $T_1 \in LM_1'(A)$ and $R \in (A_0)^{A},$ and it is a multiplier if and only if

$$\begin{aligned} R(x_1y_1) = x_1T_1(y_1) - T_1(x_1y_1) \end{aligned}$$

(12)

for any $x_1, y_1 \in A_1$.

Proof

As stated above T is written as (11). Let $x = x_1+x_0, y = y_1 + y_0 \in A$ with $x_1, y_1 \in A_1$ and $x_0, y_0 \in A_0$ be arbitrary, then we have

$$\begin{aligned} xT(y) = x_1(T_1(y) + R(y)) = x_1T_1(y) = x_1T_1(y_1) \end{aligned}$$

(13)

because $R(A) \subseteq A_0$ and $T_1(A_0) = \{0\}$. The last term in (13) is also equal to $T_1(x_1)y_1 = T(x)y$. Hence, T is a multiplier if and only if it is equal to $T(xy) = T(x_1y_1) = T_1(x_1y_1) + R(x_1y_1)$, if and only if (12) holds. $\square$

4 Linear multipliers and matrix equation

In this section, A is a finite-dimensional algebra over K. We derive a matrix equation that characterizes a (weak) multiplier for a linear mapping on A. Suppose that A is n-dimensional with basis $E = \{e_1, e_2, \dots , e_n\}$.

Lemma 4.1

A linear mapping $T: A \rightarrow A$ is a weak multiplier if and only if

$$\begin{aligned} e_iT(e_j) = T(e_i)e_j, \end{aligned}$$

(14)

and it is a multiplier if and only if

$$\begin{aligned} T(e_ie_j) = e_iT(e_j) = T(e_i)e_j, \end{aligned}$$

(15)

for all $e_i, e_j \in E$.

Proof

The necessity of the conditions (14) and (15) is obvious. Let $x = x_1e_1 + x_2e_2+ \cdots + x_ne_n, y = y_1e_1+y_2e_2+\cdots +y_ne_n \in A$ with $x_1,x_2,\dots ,x_n, y_1,y_2,\dots ,y_n \in K$. If T satisfies (14), then we have

$$\begin{aligned} xT(y)= & {} \left( \sum _i x_ie_i\right) T\left( \sum _j y_je_j\right) = \left( \sum _i x_ie_i\right) \left( \sum _j y_jT(e_j)\right) \\= & {} \sum _{i,j} x_iy_je_iT(e_j) =\sum _{i,j}x_iy_jT(e_i)e_j\\= & {} \left( \sum _ix_iT(e_i)\right) \left( \sum _jy_je_j\right) = T(x)y. \end{aligned}$$

Hence, T is a weak multiplier. Moreover, if T satisfies (15), it is a multiplier in a similar manner. $\square$

Let $\varvec{A}$ (we use the bold character) represent the multiplication table of A on E. $\varvec{A}$ is a matrix whose elements are drawn from A and given by

$$\begin{aligned} \varvec{A} = \varvec{E}^t\varvec{E}, \end{aligned}$$

(16)

where $\varvec{E} = (e_1, e_2, \dots , e_n)$ (we again use the boldface $\varvec{E}$) is the row vector consisting the basis elements and $\varvec{E}^t$ is its transpose. For a linear mapping T on A and a matrix $\varvec{B}$ over A, $T(\varvec{B})$ denotes the matrix obtained by applying T element-wise, that is, the (i, j)-element of $T(\varvec{B})$ is $T(b_{ij})$ for the (i, j)-element $b_{ij}$ of $\varvec{B}$.³ We employ the same symbol T for the representation matrix of T on E, that is,

$$\begin{aligned} T(\varvec{E}) = \varvec{E}T. \end{aligned}$$

(17)

Theorem 4.2

A linear mapping T is a weak multiplier of A if and only if

$$\begin{aligned} \varvec{A}T = T^t\varvec{A}, \end{aligned}$$

(18)

and T is a multiplier if and only if

$$\begin{aligned} T(\varvec{A}) = \varvec{A}T = T^t\varvec{A}. \end{aligned}$$

(19)

Proof

By (16) and (17) we have

$$\begin{aligned} (e_1, e_2, \dots , e_n)^t\left( T(e_1), T(e_2), \dots , T(e_n)\right) = \varvec{E}^tT(\varvec{E}) = \varvec{E}^t\varvec{E}T = \varvec{A}T \end{aligned}$$

(20)

and

$$\begin{aligned} \left( T(e_1), T(e_2), \dots , T(e_2)\right) ^t(e_1, e_2, \dots , e_n) = T(\varvec{E})^t\varvec{E} = T^t\varvec{E}^t\varvec{E} = T^t\varvec{A}. \end{aligned}$$

(21)

By Lemma 4.1, T is a weak multiplier if and only if (20) and (21) are equal, if and only if (18) holds. Moreover, T is multiplier if and only if the leftmost sides of (20) and (21) are equal to $(T(e_ie_j))_{i,j=1,2,\dots ,n} = T(\varvec{A})$, if and only if (19) holds. $\square$

The multiplication table of the opposite algebra $A^\textrm{op}$ of A is the transpose $\varvec{A}^\textrm{t}$ of $\varvec{A}$. So, the algebras with multiplication tables transposed to each other share the same (weak) multipliers.

5 Associative algebras

In this section, A is an associative algebra over K.

Proposition 5.1

If $A_0 = \{0\}$, then we have

$$\begin{aligned} M(A) = M'(A) = LM(A) = LM'(A). \end{aligned}$$

Proof

Let $T \in M'(A)$, then we have

$$\begin{aligned} T(xy)z = xyT(z) = xT(y)z\ \ \textrm{and}\ \ zT(xy) = T(z)xy = zT(x)y \end{aligned}$$

for any $x, y, z \in A$. It follows that

$$\begin{aligned} T(xy) - xT(y) \in \textrm{Ann}_l(A) \cap \textrm{Ann}_r(A) = A_0 =\{0\}. \end{aligned}$$

Hence, $T(xy) = xT(y)$ and we see $T \in M(A)$. Moreover, $T \in LM(A)$ by Proposition 2.2. $\square$

Let $a \in A$. If $xay = axy$ (resp. $xay = xya$) for any $x, y \in A$, a is called a left (resp. right) central element, and we simply call it a central element if $ax = xa$ for any $x \in A$. Let $Z_l(A)$, (resp. $Z_r(A)$, Z(A)) denote the set of all left central (resp. right central, central) elements.

For $a\in A$, $l_a$ (resp. $r_a$) denotes the left (resp. right) multiplication by a, that is,

$$\begin{aligned} l_a(x) = ax,\ \ \ r_a(x) = xa \end{aligned}$$

for $x \in A$. They are linear mappings.

Lemma 5.2

For $a \in A$ the following statements are equivalent.

(i) $l_a$ (resp. $r_a$) is a multiplier,

(ii) $l_a$ (resp. $r_a$) is a weak multiplier,

(iii) a is left (resp. right) central.

Proof

If $l_a$ is a weak multiplier, then

$$\begin{aligned} xay = xl_a(y) = l_a(x)y = axy \end{aligned}$$

for any $x, y \in A$, which implies that a is left central. Conversely, if a is left central, $l_a$ is a weak multiplier also by the above equalities. Moreover, $l_a$ is a multiplier because $l_a(xy) = axy = l_a(x)y$. The other case is analogous, and we see that these three statements are equivalent. $\square$

As can be easily proved, $Z_l(A)$ (resp. $Z_r(A)$) is a subalgebra of A containing $\textrm{Ann}_l(A)$ (resp. $\textrm{Ann}_r(A)$). Hence, we can form the quotient algebras $\bar{Z}_l(A) = Z_l(A)/ \textrm{Ann}_l(A)$ and $\bar{Z}_r(A) = Z_r(A)/\textrm{Ann}_r(A)$.

Theorem 5.3

Suppose that A has a left (resp. right) identity e. Then, any weak multiplier is a left (resp. right) multiplication by a left (resp. right) central element and it is a linear multiplier. The mapping $\phi \hspace{-.6mm}: Z_l(A)\ (resp.\ Z_r(A)) \rightarrow M'(A)= LM(A)$ sending $a \in Z_l(A)\ (resp.\ Z_r(A))$ to $l_a$ (resp. $l_r$) induces an isomorphism $\bar{\phi }\hspace{-.6mm}: \overline{Z_l}(A)\ (resp.\ \overline{Z_r}(A)) \rightarrow M(A)$ of algebras. In particular, if A is unital, M(A) is isomorphic to Z(A).

Proof

Suppose that A has a left identity e. Let $T \in M'(A)$ and set $a = T(e)$. Then we have

$$\begin{aligned} T(x) = eT(x) = T(e)x = ax \end{aligned}$$

for any $x \in A$. Hence, $T = l_a$, where $a \in Z_l(A)$ and T is a linear multiplier by Lemma 5.2. Therefore, $M'(A) = LM(A)$ and $\phi$ is surjective. Moreover, for $a \in Z_l(A)$, $\phi (a) = 0$ if and only if $ax = 0$ for any $x \in A$, if and only if $a \in \textrm{Ann}_l(A)$. Thus we have Ker$(\phi ) = \textrm{Ann}_l(A)$, and $\phi$ induces the desired isomorphism. Similarly, if A has a right identity, M(A) is isomorphic to $\overline{Z_r}(A)$. Lastly, if A has the identity, then $Z_\ell (A) = Z(A)$ and Ann$_l(A) = \{0\}$, and hence M(A) is isomorphic to Z(A). $\square$

6 3-dimensional associative algebras

Over an algebraically closed field K of characteristic not equal to 2, we have, up to isomorphism, 24 families of associative algebras of dimension 3. They consist of 5 unital algebras $U_0, U_1, U_2$, $U_3, U_4$ defined on basis $E = \{e,f,g\}$ by the multiplication tables

$$\begin{aligned} \begin{pmatrix} e&{} \quad f&{} \quad g\\ f&{} \quad 0&{} \quad 0\\ g&{} \quad 0&{} \quad 0 \end{pmatrix}, \ \begin{pmatrix} e&{} \quad f&{} \quad g\\ f&{} \quad 0&{} \quad f\\ g&{} \quad -f&{} \quad e \end{pmatrix}, \ \begin{pmatrix} e&{} \quad 0&{} \quad 0\\ 0&{} \quad f&{} \quad 0\\ 0&{} \quad 0&{} \quad g \end{pmatrix}, \ \begin{pmatrix} e&{} \quad 0&{} \quad 0\\ 0&{} \quad f&{} \quad g\\ 0&{} \quad g&{} \quad 0 \end{pmatrix}, \ \begin{pmatrix} e&{} \quad f&{} \quad g\\ f&{} \quad g&{} \quad 0\\ g&{} \quad 0&{} \quad 0 \end{pmatrix}, \end{aligned}$$

5 curled algebras $C_0, C_1, C_2, C_3, C_4$ defined by

$$\begin{aligned}{ \begin{pmatrix} 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \end{pmatrix}, \ \begin{pmatrix} 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad e \\ 0 &{} \quad -e &{} \quad 0 \end{pmatrix}, \ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {e} &{} \quad {f} &{} \quad {0} \\ {0} &{} \quad {g} &{} \quad {0} \end{pmatrix}, \ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \\ {e} &{} \quad {f} &{} \quad {g} \end{pmatrix}, \ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {e} \\ {0} &{} \quad {0} &{} \quad {f} \\ {0} &{} \quad {0} &{} \quad {g} \end{pmatrix},} \end{aligned}$$

non-unital 4 straight algebras $S_1, S_2, S_3, S_4$ defined by

$$\begin{aligned} {\begin{pmatrix} {f}\ {} &{} \quad {g} &{} \quad {0} \\ {g}\ {} &{} \quad {0} &{} \quad {0} \\ {0}\ {} &{} \quad {0} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {e}\ {} &{} \quad {0} &{} \quad {0} \\ {0}\ {} &{} \quad {g} &{} \quad {0} \\ {0}\ {} &{} \quad {0} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {e}\ {} &{} \quad {0} &{} \quad {0} \\ {0}\ {} &{} \quad {f} &{} \quad {0} \\ {0}\ {} &{} \quad {0} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {e}\ {} &{} \quad {f} &{} \quad {0} \\ {f}\ {} &{} \quad {0} &{} \quad {0} \\ {0}\ {} &{} \quad {0} &{} \quad {0} \end{pmatrix},} \end{aligned}$$

and non-unital 10 families of waved algebras $W_1$, $W_2$, $W_4$, $W_5$, $W_6$, $W_7$, $W_8$, $W_9$, $W_{10}$ and $\big \{W_3(k)\big \}_{k \in H}$ defined by

$$\begin{aligned} {\begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {e} \end{pmatrix},\ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {e} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {e} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {f} &{} \quad {g} \end{pmatrix},\ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {f} \\ {0} &{} \quad {0} &{} \quad {g} \end{pmatrix},} \end{aligned}$$

$$\begin{aligned} { \begin{pmatrix} e &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad f &{} \quad g \end{pmatrix},\ \begin{pmatrix} {e} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {0} &{} \quad {f} \\ {0} &{} \quad {0} &{} \quad {g} \end{pmatrix},\ \begin{pmatrix} {0} &{} \quad {e} &{} \quad {0} \\ {e} &{} \quad {f} &{} \quad {0} \\ {0} &{} \quad {g} &{} \quad {0} \end{pmatrix},\ \begin{pmatrix} {0} &{} \quad {e} &{} \quad {0} \\ {e} &{} \quad {f} &{} \quad {g} \\ {0} &{} \quad {0} &{} \quad {0} \end{pmatrix} \ \text{ and }\ \begin{pmatrix} {0} &{} \quad {0} &{} \quad {0} \\ {0} &{} \quad {e} &{} \quad {0} \\ {0} &{} \quad k{e} &{} \quad {e} \end{pmatrix},} \end{aligned}$$

respectively, where H is a subset of K such that $K = H\cup -H$ and $H\cap -H = \{0\}$ (see [3] for details). We determine the (weak) multipliers of them below.

(0) $A = C_0$ is the zero algebra, so by Proposition 2.5, we have

$$\begin{aligned} M'(A) = {A}^{A}, \ \,M(A) = \{T \in {A}^{A}\,|\, T(0) = 0 \}\ \text{ and }\ \, LM(A) = LM'(A) = L(A). \end{aligned}$$

(i) The unital algebras $U_0, U_2$, $U_3, U_4$ are commutative, so for such A we have

$$\begin{aligned} M(A) = LM(A) = M'(A) = LM'(A) = \{ l_x | x \in A \} \cong A \end{aligned}$$

by Theorem 5.3. For $A = U_1$, we have

$$\begin{aligned} M(A) = LM(A) = M'(A) = LM'(A) \cong Z(A) = Ke. \end{aligned}$$

(ii) For $A = C_1$, we observe that $A_0 = \text{ Ann}_l(A) = \text{ Ann}_r(A) =Ke$, and we have a nihil decomposition $A = A_1 \oplus A_0$ with $A_1 = Kf + Kg$. Let $T_1 \in M'_1(A)$, then by Theorem 3.1, $T_1$ is a linear mapping such that $T_1(Ke) = \{0\}$. Let

$$\begin{aligned} T_1 = \begin{pmatrix}0&{} \quad 0&{} \quad 0\\ 0&{} \quad q&{} \quad r\\ 0&{} \quad t&{} \quad u\end{pmatrix} \end{aligned}$$

(22)

with $q,r,t,u \in K$ be the representation matrix of $T_1$ on E. By Theorem 4.2, $T_1$ is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix} 0&{} \quad 0&{} \quad 0\\ 0&{} \quad te&{} \quad ue\\ 0&{} \quad -qe&{} \quad -re \end{pmatrix} = \varvec{A}T_1 = T_1^\textrm{t}\varvec{A} = \begin{pmatrix} 0&{} \quad 0&{} \quad 0\\ 0&{} \quad -te&{} \quad qe\\ 0&{} \quad -ue&{} \quad re \end{pmatrix}, \end{aligned}$$

if and only if $r = t = 0$ and $q = u$. Hence, $M'_1(A) = \{T_q\,\big |\,q \in K\}$, where $T_q = {\begin{pmatrix} 0&{} \quad 0&{} \quad 0\\ 0&{} \quad q&{} \quad 0\\ 0&{} \quad 0&{} \quad q \end{pmatrix}}$. By Theorem 3.1 we see

$$\begin{aligned} M'(A) = \{T_q\,\big |\,q \in K\} \oplus (Ke)^A \end{aligned}$$

and

$$\begin{aligned} LM'(A) = \left\{ {\begin{pmatrix} a&{} \quad b&{} \quad c\\ 0&{} \quad q&{} \quad 0\\ 0&{} \quad 0&{} \quad q \end{pmatrix}}\, \Big |\,a,b,c,q \in K\right\} . \end{aligned}$$

By examining the multiplication table of A, we find that  $\alpha \beta = (xv-yz)e$   for $\alpha = xf+yg, \beta = zf+vg \in A_1$ with $x,y,z,v \in K.$ By Corollary 3.2, $T \in M'(A)$ is given by $T = T_q + R$ with $R \in (Ke)^A$ and this T is a multiplier if and only if

$$\begin{aligned} R((xv-yz)e)= & {} R(\alpha \beta )\ =\ \alpha T_q(\beta ) - T_q(\alpha \beta )\\ {}= & {} \alpha (q\beta ) - T_q((xv-yz)e)\ =\ q(xv-yz)e \end{aligned}$$

for any $\alpha$ and $\beta$, if and only if $R(xe) = qxe$ for all $x \in K$. Let $S_q$ be the scalar multiplication in A by $q \in K$. Then, we see

$$\begin{aligned} (T-S_q)(A) = (T_q - S_q + R)(A) \subseteq A_0 = Ke \end{aligned}$$

and

$$\begin{aligned} (T-S_q)(xe) = T_q(xe) + R(xe) - S_q(xe) = 0 + qxe -qxe = 0 \end{aligned}$$

for any $x \in K$, that is, $(T - S_q)(Ke)= \{0\}$. Thus, we conclude

$$\begin{aligned} M(A) = \{S_q\,\big |\,q \in K\} \oplus \big \{R \in (Ke)^A\,\big |\,R(Ke) = \{0\}\big \}, \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ {\begin{pmatrix} a&{} \quad b&{} \quad c\\ 0&{} \quad a&{} \quad 0\\ 0&{} \quad 0&{} \quad a \end{pmatrix}}\, \Big |\,a,b,c \in K\right\} . \end{aligned}$$

(iii) $A = C_2$: Because Ann$_l(A) = Ke$ and Ann$_r(A) = Kg$, we see $A_0 = \{0\}$. Hence, any weak multiplier T is a linear multiplier by Proposition 5.1. By Theorem 4.2,

$$\begin{aligned} T =\begin{pmatrix} a&{} \quad b&{} \quad c\\ p&{} \quad q&{} \quad r\\ s&{} \quad t&{} \quad u \end{pmatrix} \end{aligned}$$

(23)

with $a,b,c,p,q,r,s,t,u \in K$ is a (weak) multiplier if and only if

$$\begin{aligned} \begin{pmatrix} 0&{} \quad 0&{} \quad 0\\ ae+pf&{} \quad be+qf&{} \quad ce+rf\\ pg&{} \quad qg&{} \quad rg \end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix} pe&{} \quad pf+sg&{} \quad 0\\ qe&{} \quad qf+tg&{} \quad 0\\ re&{} \quad rf+ug&{} \quad 0 \end{pmatrix}, \end{aligned}$$

if and only if $b = c = p = r = s = t = 0$ and $a = q = u$, that is, T is the scalar multiplication $S_a$ by a. Consequently,

$$\begin{aligned} M(A) = M'(A) = LM(A) = LM'(A) = \{S_a\,\big |\,a \in K\} \cong K. \end{aligned}$$

(iv) $C_3$ and $C_4$ are opposite to each other, and share the same (weak) multipliers by Proposition 2.6. Let $A = C_3$, then, A has a left identity g, $Z_l(A) = A$ and Ann$_l(A) = Ke+Kf$. Hence, by Theorem 5.3,

$$\begin{aligned} M(A) = M'(A) = LM(A) = LM'(A) = A/(Ke+Kg) = \{S_a\,\big |\,a \in K\}. \end{aligned}$$

(v) $A = S_1$: We have $A_0 = \text{ Ann}_l(A) = \text{ Ann}_r(A) = Kg$, and $A = A_1 \oplus A_0$ with $A_1 = Ke + Kf$. Then, $T_1 \in M'_1(A)$ is a linear mapping with $T(Kg) = \{0\}$. Let

$$\begin{aligned} T_1 = \begin{pmatrix} a&{} \quad b&{} \quad 0\\ p&{} \quad q&{} \quad 0\\ 0&{} \quad 0&{} \quad 0 \end{pmatrix} \end{aligned}$$

(24)

with $a,b,p,q \in K$ be its representation on E. $T_1$ is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix} af+pg&{} \quad bf+qg&{} \quad 0\\ ag&{} \quad bg&{} \quad 0\\ 0&{} \quad 0&{} \quad 0 \end{pmatrix} = \varvec{A}T_1 = T_1^\textrm{t}\varvec{A} = \begin{pmatrix} af+pg&{} \quad ag&{} \quad 0\\ bf+qg&{} \quad bg&{} \quad 0\\ 0&{} \quad 0&{} \quad 0 \end{pmatrix}, \end{aligned}$$

if and only if $b = 0$ and $a = q$. Hence,

$$\begin{aligned} M'(A) = \left\{ T_1^{a,p}\,|\,a,p \in K\right\} \oplus (Kg)^A\ \, \text{ with } \ \,T_1^{a,p} = \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ p&{} \quad a&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix}}. \end{aligned}$$

So, $T \in M'(A)$ is written as $T = T_1^{a,p} + R$ with $R \in (Kg)^A$, and this T is multiplier if and only if

$$\begin{aligned} R(xzf+(xv+yz)g)= & {} R(\alpha \beta )\ =\ \alpha T_1^{a,p}(\beta ) - T_1^{a,p}(\alpha \beta )\\= & {} (xe+yf)(aze+(pz+av)f) - axzf\\= & {} (pxz+a(xv+yz))g \end{aligned}$$

for any $\alpha = xe+yf, \beta = ze+vf \in A_1$ with $x,y,z,v \in K$, if and only if  $R(xf+yg) = (px+ay)g$  for all $x, y \in K$. Let $T^{a,p} = \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ p&{} \quad a&{} \quad 0\\ 0&{} \quad p&{} \quad a\end{pmatrix}}$, then $(T-T^{a,p})(A) \subseteq Kg$, and

$$\begin{aligned} (T-T^{a,p})(xf+yg)= & {} \left( T_1^{a,p}+R-T^{a,p}\right) (xf+yg)\\= & {} axf + (px+ay)g - (axf+pxg+ayg) = 0 \end{aligned}$$

for any $x, y \in K$. Thus, $(T - T^{a,p})(Kf+Kg) = \{0\}$, and hence

$$\begin{aligned} M(A) = \{T^{a,p}\,|\,a, p \in K\} \oplus \big \{R \in (Kg)^A\,\big |\,R(Kf+Kg) = \{0\}\big \}. \end{aligned}$$

By intersecting $M'(A)$ and M(A) with L(A), we obtain

$$\begin{aligned} LM'(A)= & {} \left\{ {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ p&{} \quad a&{} \quad 0\\ s&{} \quad t&{} \quad u\end{pmatrix}}\,\Big |\,a,p,s,t,u \in K\right\} \ \\{} & {} \text{ and }\ \, LM(A) = \left\{ {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ p&{} \quad a&{} \quad 0\\ s&{} \quad p&{} \quad a\end{pmatrix}}\,\Big |\,a,p,s \in K\right\} . \end{aligned}$$

(vi) $A = S_2$: We have $A_0 = \text{ Ann}_l(A) = \text{ Ann}_r(A) = Kg$, and $A = A_1 \oplus A_0$ with $A_1 = Ke + Kf$. Let a linear mapping $T_1 \in M'_1(A)$ be represented as (24), then $T_1$ is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix}ae&{} \quad be&{} \quad 0\\ pg&{} \quad qg&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix}ae&{} \quad pg&{} \quad 0\\ be&{} \quad qg&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix}, \end{aligned}$$

if and only if $b = p = 0$. Hence,

$$\begin{aligned} M'(A) = \{T_1^{a,q}\,\big |\,a,q \in K\}\oplus (Kg)^A\ \,\text{ with }\ \, T_1^{a,q} = \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad q&{} \quad 0\\ 0&{} \quad 0&{} \quad 0\end{pmatrix}}, \end{aligned}$$

and

$$\begin{aligned} LM'(A) = \left\{ \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad q&{} \quad 0\\ s&{} \quad t&{} \quad u\end{pmatrix}}\,\Big |\,a,q,s,t,u \in K\right\} . \end{aligned}$$

By Corollary 3.2, a weak multiplier T written as $T = T_1^{a,q} + R$ for $a,q \in K$ and $R \in (Kg)^A$ is multiplier if and only if

$$\begin{aligned} R(xze+yvg)= & {} R(\alpha \beta )\ =\ \alpha T_1^{a,q}(\beta ) -T_1^{a,q}(xze+yvg)\\= & {} (xe+yf)(aze+qvf) - axze \ =\ yqvg \end{aligned}$$

for any $\alpha = xe+yf, \beta = ze+vf \in A_1$ with $x,y,z,v \in K$, if and only if $R(xe+yg) = qyg$ for all $x, y \in K$. Let $T^{a,q} = \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad q&{} \quad 0\\ 0&{} \quad 0&{} \quad q\end{pmatrix}}$, then we have $(T-T^{a,q})(xe+yg) = 0$ for any $x, y \in K$, following the same calculation as in (v) above. Hence, $(T-T^{a,q})(Ke+Kg) = \{0\}$, and we have

$$\begin{aligned} M(A) = \left\{ T^{a,p}\,|\,a,p \in K\right\} \oplus \big \{R \in (Kg)^A\,\big |\,R(Ke+Kg) = \{0\}\big \} \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad p&{} \quad 0\\ 0&{} \quad t&{} \quad 0\end{pmatrix}}\,\Big |\,a,p,t \in K\right\} . \end{aligned}$$

(vii) $A = S_3$: We have $A_0 = Kg$ and $A = A_1 \oplus A_0$ with $A_1 = Ke+Kf$. As $A_1$ is a subalgebra of A, by Theorem 3.1 we obtain the equalities (10) in Sect. 3. Because $A_1$ is a commutative unital algebra,

$$\begin{aligned} M(A_1) = M'(A_1) = A_1 = \left\{ \small {\begin{pmatrix}a&{} \quad 0\\ 0&{} \quad b\end{pmatrix}}\,\Big |\,a, b \in K\right\} \end{aligned}$$

by Theorem 5.3. Note that $A^2 = Ke+Kf$. Hence,

$$\begin{aligned} M'(A) = A_1 \oplus (Kg)^A\ \,\text{ and }\ \, M(A) = A_1 \oplus \{T \in (Kg)^A\,|\,T(Ke+Kf) = 0\}. \end{aligned}$$

Intersecting with L(A) we have

$$\begin{aligned} LM'(A)= & {} \left\{ \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad b&{} \quad 0\\ s&{} \quad t&{} \quad u\end{pmatrix}}\, \Big |\,a,b,s,t,u \in K\right\} \ \\{} & {} \textrm{and}\ \ LM(A) = \left\{ \small {\begin{pmatrix}a&{} \quad 0&{} \quad 0\\ 0&{} \quad b&{} \quad 0\\ 0&{}0&{}u\end{pmatrix}}\,\Big |\,a,b,u \in K\right\} . \end{aligned}$$

(viii) $A = S_4$: We have $A = A_1 \oplus A_0$ with $A_0 = Kg$ and $A_1 = Ke+Kf$. Because $A_1$ is a commutative unital subalgebra of A, similarly to the previous case, we obtain

$$\begin{aligned} M'(A) = A_1 \oplus (Kg)^A = \left\{ \small {\begin{pmatrix}a&{}0\\ b&{}a\end{pmatrix}}\,\Big |\,a, b \in K\right\} \oplus (Kg)^A, \end{aligned}$$

$$\begin{aligned}{} & {} M(A)=A_1 \oplus \big \{T \in (Kg)^A\,\big |\,T(Ke+Kf) = \{0\}\big \}, \\{} & {} LM'(A) \!=\! \left\{ \small {\begin{pmatrix}a&{}0&{}0\\ b&{}a&{}0\\ s&{}t&{}u\end{pmatrix}}\,\Big |\,a,b,s,t,u \!\in \! K\right\} \ \textrm{and}\ \ \!LM(A) \!=\! \left\{ \small {\begin{pmatrix}a&{}0&{}0\\ b&{}a&{}0\\ 0&{}0&{}u\end{pmatrix}}\,\Big |\,a,b,u \!\in \! K\right\} . \end{aligned}$$

(ix) $A = W_1$: We have $A = A_1 \oplus A_0$ with $A_0 = Ke+Kf$ and $A_1 = Kg$. Let $T_1 \in M'_1(A)$, then $T_1$ is a linear mapping with $T_1(A_0) = \{0\}$. So $T_1$ is determined by $T_1(g) = ag$ with $a \in K$. Denoting this $T_1$ as $T^a_1$, we have

$$\begin{aligned} M'(A) = \{T^a_1\,|\,a \in K\} \oplus (Ke+Kf)^A. \end{aligned}$$

A weak multiplier $T = T^a_1 + R$ with $R \in (Ke+Kf)^A$ is a multiplier if and only if

$$\begin{aligned} R(xye) = R((xg)(yg)) = xgT^a_1(yg) - T^a_1(xye) =axye \end{aligned}$$

for all $x, y \in K$, if and only if $R(xe) = axe$ for any $x\in K$. Let $T_a = \small {\begin{pmatrix}a&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad a\end{pmatrix}}$. Then, $(T-T_a)(Ke) = \{0\}$ and it follows that

$$\begin{aligned} M(A) = \{T_a\,\big |\,a\in K\} \oplus \big \{R \in (Ke+Kf)^A\,\big |\, R(Ke) = \{0\}\big \}. \end{aligned}$$

Also we have

$$\begin{aligned} LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ p&{}\quad q&{}\quad r\\ 0&{}\quad 0&{}\quad u\end{pmatrix}}\,\Big |\,a,b,c,p,q,r,u \in K\right\} \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad q&{}\quad r\\ 0&{}\quad 0&{}\quad a\end{pmatrix}}\,\Big |\,a,b,c,q,r \in K\right\} . \end{aligned}$$

(x) $A = W_2$⁴: We have $A = A_1 \oplus A_0$ with $A_0 = Ke$ and $A_1 = Kf+Kg$. Let $T \in M'_1(A)$, then T is a linear mapping with $T(Ke) = \{0\}$ and can be represented as (22). Then, T is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad qe&{}\quad re\end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad te&{}\quad 0\\ 0&{}\quad ue&{}\quad 0\end{pmatrix}, \end{aligned}$$

if and only if $r = t = 0$ and $q = u$. Hence,

$$\begin{aligned} M'(A) = \{T_q\,|\,q \in K\}\oplus (Ke)^A\ \,\text{ with }\ \, T_q = \small {\begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}. \end{aligned}$$

So,

$$\begin{aligned} LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}\,\Big |\,a,b,c,q \in K\right\} . \end{aligned}$$

A weak multiplier $T = T_q+R$ with $R \in (Ke)^A$ is a multiplier if and only if

$$\begin{aligned} R(yze) = R(\alpha \beta ) = \alpha T_q(\beta ) - T_q(yze) = \alpha (q\beta ) = qyze \end{aligned}$$

for any $\alpha = xf+yg, \beta = zf+vg \in A_1$ with $x,y,z,v \in K$, if and only if $R(xe) = qxe$ for all $x \in K$. Let $S_a$ be the scalar multiplication by $a \in K$. Then, $(T-S_a)(Ke) = \{0\}$, and hence,

$$\begin{aligned} M(A) = \{S_a\,|\,a \in K\} \oplus \big \{R \in (Ke)^A\,\big |\,R(Ke) = \{0\}\big \} \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad a&{}\quad 0\\ 0&{}\quad 0&{}\quad a\end{pmatrix}}\,\Big |\,a,b,c \in K\right\} . \end{aligned}$$

(xi) $A = W_4$: We have $A = A_1 \oplus A_0$ with $A_0 = Kf + Kg$ and $A_1 = Ke$. Because $A_1$ is a subalgebra isomorphic to the base field K, with the scalar multiplication $S_1^a$ by $a\in K$, we see

$$\begin{aligned} M'(A) = \{S_1^a\,|\,a \in K\} \oplus (fK+gK)^A \end{aligned}$$

and

$$\begin{aligned} M(A) = \{S_1^a\,|\,a \in K\} \oplus \big \{R \in (fK+gK)^A\,\big |\,R(Ke) = \{0\}\big \} \end{aligned}$$

by Theorem 3.1. Taking the intersections with L(A) we have

$$\begin{aligned} LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad 0&{}\quad 0\\ p&{}\quad q&{}\quad r\\ s&{}\quad t&{}\quad u\end{pmatrix}}\,\Big |\,a,p,q,r,s,t,u \in K\right\} \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad r\\ 0&{}\quad t&{}\quad u\end{pmatrix}}\,\Big |\,a,q,r,t,u \in K\right\} . \end{aligned}$$

(xii) $W_5$ and $W_6$ are opposite. Let $A = W_5$, then $A = A_1\oplus A_0$ with $A_0 = Ke$ and $A_1 = Kf+Kg$. Since $A_1$ is a subalgebra of A and has a left identity g, we have

$$\begin{aligned} M(A_1) = LM(A_1) = M'(A_1) = LM'(A_1) \cong (A_1)/Kf \cong Kg \end{aligned}$$

by Theorem 5.3. So, any element in $M(A_1)$ is a scalar multiplication $S_1^q$ in $A_1$ by $q \in K$. By Theorem 3.1 we have

$$\begin{aligned}{} & {} M'(A) = \{S_1^q\,|\,q \in K\} \oplus (Ke)^A, \\{} & {} M(A) = \{S_1^q\,|\,q \in K\} \oplus \big \{R \in (Ke)^A\,\big |\,R(Kf+Kg) = \{0\}\big \}, \\{} & {} LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}\,\Big |\,a,b,c,q \in K\right\} \ \, \textrm{and}\ \, LM(A) = \left\{ \small {\begin{pmatrix}a&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}\,\Big |\,a,q \in K\right\} . \end{aligned}$$

(xiii) $W_7$ and $W_8$ are opposite. Let $A = W_7$, then we see $A_0 = \text{ Ann}_r(A) = \{0\}$. Hence, any weak multiplier is a linear multiplier by Proposition 5.1, and a linear mapping T represented as (23) is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix}ae&{}\quad be&{}\quad ce\\ 0&{}\quad 0&{}\quad 0\\ pf+sg&{}\quad qf+tg&{}rf+ug\end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix}ae&{}\quad sf&{}\quad sg\\ be&{}\quad tf&{}\quad tg\\ ce&{}\quad uf&{}\quad ug\end{pmatrix}, \end{aligned}$$

if and only if $b = c = p = r = s = t = 0$ and $q = u$. Therefore,

$$\begin{aligned} M(A) = LM(A) = M'(A) = LM'(A) = LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}\,\Big |\,a,q \in K\right\} . \end{aligned}$$

(xiv) $W_9$ and $W_{10}$ are opposite. Let $A = W_9$. Then, because $A_0 = \text{ Ann}_l(A) = \{0\}$, any weak multiplier is a linear multiplier and a linear mapping T represented as (23) is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix}pe&{}\quad qe&{}\quad re\\ ae+pf&{}\quad be+qf&{}\quad ce+rf\\ pg&{}\quad qg&{}\quad rg\end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix}pe&{}\quad ae+pf+sg&{}\quad 0\\ qe&{}\quad be+qf+tg&{}\quad 0\\ re&{}\quad ce+rf+ug&{}\quad 0\end{pmatrix}, \end{aligned}$$

if and only if $c = p = r = s = t = 0, a = q = u$. Therefore,

$$\begin{aligned} LM(A) = M(A) = LM'(A) = M'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad 0\\ 0&{}\quad a&{}\quad 0\\ 0&{}\quad 0&{}\quad a\end{pmatrix}}\,\Big |\,a,b \in K\right\} . \end{aligned}$$

(xv) $A = W_3(k)$: We have $A = A_1 \oplus A_0$ with $A_0 = Ke$ and $A_1 = Kf+Kg$. Then, $T \in M'_1(A)$ is a linear mapping with $T(Ke) = \{0\}$ represented as (22). It is a weak multiplier if and only if

$$\begin{aligned} \begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad qe&{}\quad re\\ 0&{}\quad (kq+t)e&{}\quad (kr+u)e\end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad (q+kt)e&{}\quad te\\ 0&{}\quad (r+ku)e&{}\quad ue\end{pmatrix}. \end{aligned}$$

(25)

When $k = 0$, (25) holds if and only if $r = t$, and otherwise it holds if and only if $r = t = 0$ and $q =u$. Thus,

$$\begin{aligned} M'(A) = \{T_1^{q,r,u}\,\big |\,q,r,u \in K\} \oplus (Ke)^A,\ LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad q&{}r\\ 0&{}\quad r&{}\quad u\end{pmatrix}}\,\Big |\,a,b,c,q,r,u \in K\right\} \end{aligned}$$

when $k = 0$, and

$$\begin{aligned} M'(A) = \{T_1^{q}\,\big |\,q \in K\} \oplus (Ke)^A,\ \, LM'(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}\,\Big |\,a,b,c,q \in K\right\} \end{aligned}$$

when $k \ne 0$, where

$$\begin{aligned} T_1^{q,r,u} = \small {\begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad r\\ 0&{}\quad r&{}\quad u\end{pmatrix}}\ \, \textrm{and}\ \, T_1^{q} = \small {\begin{pmatrix}0&{}\quad 0&{}\quad 0\\ 0&{}\quad q&{}\quad 0\\ 0&{}\quad 0&{}\quad q\end{pmatrix}}. \end{aligned}$$

Now, when $k = 0$, $T = T_1^{q,r,u} + R$ with $R \in (Ke)^A$ is multiplier if and only if

$$\begin{aligned} R((xz+yv)e)= & {} R(\alpha \beta )\ =\ \alpha T_1^{q,r,u}(\beta ) - T_1^{q,r,u}((xz+yv)e)\\= & {} \alpha ((qz+rv)f+(rz+uv)g) = (qxz+uyv+r(xv+yz))e \end{aligned}$$

for any $\alpha = xf+yg, \beta = zf+vg \in A_1$ with $x,y,z,v \in K$, if and only if $q = u, r = 0$ and $R(xe) = qxe$ for all $x \in K$. While, when $k \ne 0$, $T = T_1^q + R$ with $R \in (Ke)^A$ is a multiplier if and only if

$$\begin{aligned} R((xz + y(kz+v))e)= & {} R(\alpha \beta )\ =\ \alpha T_1^q(\beta ) - T_1^q(xue+y(kz+v)e)\\= & {} \alpha (q\beta )\ =\ q(xz + y(kz+v))e \end{aligned}$$

for any $\alpha , \beta$, if and only if $R(xe) = qxe$ for any $x \in K$. In both cases, with the scalar multiplication $S_a$ by $a \in K$, we have  $(T-S_a)(Ke) = \{0\}$. Therefore, for arbitrary k (whether k is zero or nonzero) we conclude that

$$\begin{aligned} M(A) = \{S_a\,\big |\, a \in K\} \oplus \big \{R \in (Ke)^A\,\big |\,R(Ke) = \{0\}\big \} \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad c\\ 0&{}\quad a&{}\quad 0\\ 0&{}\quad 0&{}\quad a\end{pmatrix}}\,\Big |\,a,b,c \in K\right\} . \end{aligned}$$

7 3-dimensional zeropotent algebras

An algebra A is zeropotent if $x^2 = 0$ for all $x \in A$. A zeropotent algebra A is anti-commutative, that is, $xy = -yx$ for all $x,y \in A$. Thus we see

$$\begin{aligned} A_0 = \text{ Ann}_l(A) = \text{ Ann}_r(A). \end{aligned}$$

Let A be a zeropotent algebras of dimension 3 over a field K with char$(K) \ne 2$. Let ${E} = \{e, f, g\}$ be a basis of A. Because A is anti-commutative, the multiplication table $\varvec{A}$ of A on E is given as

$$\begin{aligned} \varvec{A} = \begin{pmatrix}0&{}\quad \alpha &{}\quad -\beta \\ alpha&{}\quad 0&{}\quad \gamma \\ \beta &{}\quad -\gamma &{}\quad 0\end{pmatrix} \ \ \text{ with }\ \ {\left\{ \begin{array}{ll} \gamma = fg = a_{11}e + a_{12}f + a_{13}g\\ \beta = ge = a_{21}e + a_{22}f + a_{23}g\\ \alpha = ef = a_{31}e + a_{32}f + a_{33}g \end{array}\right. } \end{aligned}$$

for $a_{ij} \in K$. We call $\small {\begin{pmatrix}a_{11}&{}\quad a_{12}&{}\quad a_{13}\\ a_{21}&{}\quad a_{22}&{}\quad a_{23}\\ a_{31}&{}\quad a_{32}&{}\quad a_{33}. \end{pmatrix}}$ the structural matrix of A. The rank of A is defined as the rank of its structural matrix.

Lemma 7.1

If $\textrm{rank}(A) \ge 2$, then $A_0 = \{0\}$.

Proof

If $\textrm{rank}(A) \ge 2$, at least two of $\alpha = ef, \beta = ge, \gamma = fg$ are linearly independent. Suppose that $\alpha$ and $\beta$ are linearly independent (the other cases are similar). If $x = ae +bf +cg$ with $a,b,c \in K$ is in $\text{ Ann}_l(A)$, then $xe = -b\alpha + c\beta$ and $xf = a\alpha -c\gamma$ are both zero. It follows that $a = b = c = 0$. Hence, we have $A_0 = \text{ Ann}_l(A) = \{0\}$.

$\square$

Theorem 7.2

Let A be a zeropotent algebra of dimension 3 with rank$(A) \ge 2$ over K. Then, any weak multiplier of A is the scalar multiplication $S_a$ for some $a \in K$, that is,

$$\begin{aligned} M(A) = M'(A) = LM(A) = LM'(A) = \{S_a\,|\,a \in K\}. \end{aligned}$$

Proof

By Lemma 7.1 and Corollary 2.3, any weak multiplier T is a linear mapping. Let $T \in L(A)$ be represented as (23). By Theorem 4.2, T is a weak multiplier if and only if $\varvec{A}T = T^\textrm{t} \varvec{A}$, if and only if

$$\begin{aligned} {\begin{pmatrix}p\alpha -s\beta &{}\quad q\alpha -t\beta &{}\quad r\alpha -u\beta \\ -a\alpha +s\gamma &{}\quad -b\alpha +t\gamma &{}\quad -c\alpha +u\gamma \\ a\beta -p\gamma &{}\quad b\beta -q\gamma &{}\quad c\beta -r\gamma \end{pmatrix} = \begin{pmatrix}-p\alpha +s\beta &{}\quad a\alpha -s\gamma &{}\quad -a\beta +p\gamma \\ -q\alpha +t\beta &{}\quad b\alpha -t\gamma &{}\quad -b\beta +q\gamma \\ -r\alpha +u\beta &{}\quad c\alpha -u\gamma &{}\quad -c\beta +r\gamma \end{pmatrix}}\nonumber \\ \end{aligned}$$

(26)

holds. Suppose that $\alpha , \beta$ are linearly independent (the other cases are similar). Then, by comparing the (1,1)-elements of the two matrices in (26), we have $p\alpha - s\beta = -p\alpha +s\beta$, which implies $p = s = 0$. Comparing the (1,2)-elements and (1,3)-elements, we have $q\alpha - t\beta = a\alpha -s\gamma = a\alpha$ and $r\alpha -u\beta = -a\beta +p\gamma = -a\beta$. It follows that $a = q = u$ and $r = t = 0$. Furthermore, comparing (2,2)-elements and (3,3)-elements, we see $b = c = 0$. Consequently, (26) holds if and only if  $b = c = p = r = s = t = 0\ \ \text{ and }\ \ a = q = u$, that is, $T = S_a$. $\square$

In [4] we classified the zeropotent algebras of dimension 3 over an algebraically field K of characteristic not equal to 2. Up to isomorphism, we have 10 families of zeropotent algebras. They are

$$\begin{aligned} Z_0, Z_1, Z_2, Z_3, \{Z_4(a)\}_{a\in H}, Z_5, Z_6, \{Z_7(a)\}_{a\in H}, Z_8 \text{ and } Z_9 \end{aligned}$$

defined respectively by the structural matrices

$$\begin{aligned}{ \begin{pmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{pmatrix}, \begin{pmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}, \begin{pmatrix} 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}, \begin{pmatrix} 0&{}\quad 1&{}\quad 0\\ -1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0 \end{pmatrix}, \begin{pmatrix} 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad a\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix},} \end{aligned}$$

$$\begin{aligned} \begin{pmatrix} 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}, \begin{pmatrix} 0&{}\quad 1&{}\quad 1\\ 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}, \begin{pmatrix} 1&{}\quad a&{}\quad 0\\ 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}, \begin{pmatrix} 1&{}\quad 2&{}\quad 2\\ 0&{}\quad 1&{}\quad 2\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix} \text{ and } \begin{pmatrix} 1&{}\quad 3&{}\quad 3\\ 0&{}\quad 1&{}\quad 3\\ 0&{}\quad 0&{}\quad 1 \end{pmatrix}. \end{aligned}$$

$Z_0$ is the zero algebra, and $Z_1$ is isomorphic to the 3-dimensional associative algebra $C_1$, and their (weak) multipliers are already determined in Sect. 6. The algebras $Z_3$ to $Z_9$ have rank greater or equal to 2, and they are covered by Theorem 7.2.

Thus, only $A = Z_2$ remains to be analyzed. The multiplication table $\varvec{A}$ of A is ${\begin{pmatrix} 0&{}\quad g&{}\quad 0\\ -g&{}\quad 0&{}\quad g\\ 0&{}\quad -g&{}\quad 0\end{pmatrix}}$. We see $A_0 = \text{ Ann}_l(A) = \text{ Ann}_r(A) = K(e+g),$ and we have the nihil decomposition $A = A_0 \oplus A_1$ with $A_1 = Ke + Kf$. A weak multiplier $T \in M'_1(A)$ is a linear mapping represented by ${\begin{pmatrix} a&{}\quad b&{}\quad c\\ p&{}\quad q&{}\quad r\\ 0&{}\quad 0&{}\quad 0 \end{pmatrix}}$ satisfying

$$\begin{aligned} \begin{pmatrix} pg&{}\quad qg&{}\quad rg\\ -ag&{}\quad -bg&{}\quad -cg\\ -pg&{}\quad -qg&{}\quad -rg \end{pmatrix} = \varvec{A}T = T^\textrm{t}\varvec{A} = \begin{pmatrix} -pg&{}\quad ag&{}\quad pg\\ -qg&{}\quad bg&{}\quad qg\\ -rg&{}\quad cg&{}\quad rg\end{pmatrix}. \end{aligned}$$

Hence, $a = -c= q$ and $b = p = r = 0$. Let $T_a$ be this linear mapping, then by Theorem 3.1 we have

$$\begin{aligned} M'(A) = \{T_a\,|\,a \in K\} \oplus (K(f+g))^A \end{aligned}$$

and

$$\begin{aligned} LM'(A) = \left\{ {\begin{pmatrix} a+s&{}\quad t&{}\quad -a+u\\ 0&{}\quad a&{}\quad 0\\ s&{}\quad t&{}\quad u \end{pmatrix}}\, \Big |\,a,s,t,u \in K\right\} . \end{aligned}$$

By Corollary 3.2, a weak multiplier $T = T_a + R$ with $R \in (K(e+g))^A$ becomes a multiplier if and only if for any $\zeta = xe+yf$ and $\eta = ze+vf$ with $x, y, z, v \in K$,

$$\begin{aligned} R((xv-yz)g)= & {} R(\zeta \eta )\ =\ \zeta T_a(\eta ) - T_a((xv-yz)g)\\= & {} \zeta (a\eta ) + a(xv-yz)e\ =\ a(xv-yz)(e+g) \end{aligned}$$

holds. It follows that $R(xg) = ax(e+g)$ for all $x \in K$. Let $S_a$ be the scalar multiplication by a, then $(T-S_a)(A) \subseteq K(e+g)$ and $(T - S_a)(Kg) = \{0\}$. Hence, we obtain

$$\begin{aligned} M(A) = \{S_a\,|\,a \in K\} \oplus \big \{R \in (K(e+g))^A\,\big |\, R(Kg) = \{0\}\big \}, \end{aligned}$$

and

$$\begin{aligned} LM(A) = \left\{ \small {\begin{pmatrix}a+s&{}\quad t&{}\quad 0\\ 0&{}\quad a&{}\quad 0\\ s&{}\quad t&{}\quad a\end{pmatrix}}\,\Big |\,a,s,t \in K\right\} = \left\{ \small {\begin{pmatrix}a&{}\quad b&{}\quad 0\\ 0&{}\quad c&{}\quad 0\\ a-c&{}\quad b&{}\quad c\end{pmatrix}}\,\Big |\,a,b,c \in K\right\} . \end{aligned}$$